A058067, multiplicative in a complicated way
David Wilson
davidwwilson at comcast.net
Tue Jun 14 00:00:55 CEST 2005
Thank you for your message on A059067 and A046699. With that hint, I think
I have cracked A059067, in the sense that I now have an computationally
efficient multiplicative formula.
Let p be prime. Construct sequence f_p by writing each positive number k+1
times, where k is exponent of the largest power of p dividing n. Let g_p be
the running sum of this sequence, starting with g_p(0) = 0.
For example,
f_2 = 1,2,2,3,4,4,4,5,6,6,7,8,8,8,8,9,10,10,11,12,12,12,...
g_2 = 0,1,3,5,8,12,16,20,25,31,37,...
f_2 is essentially A046699. Similarly
f_3 = 1,2,3,3,4,5,6,6,7,8,9,9,9,10,11,12,12,13,14,15,15,...
g_3 = 0,1,3,6,9,13,18,24,30,37,45,...
Then it appears that A059067 is multiplicative with a(p^e) = p^(p*g_p(e)).
I have checked that with respect to the first 100 elements of g_2, g_3 and
g_5, the multiplicative and explicit formulae for A059067 agree. Given that
and certain expected behaviors of the A059067 formula, I am fairly confident
that I have the multiplicative formula right.
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